Section 5.4: Graphing and Trigonometric Functions

In the previous section, we focused on the algebraic properties of trigonometric functions.  In this section, we will focus on their geometric properties.  Specifically, we will use our transformational graphing technique to quickly sketch the graphs of functions with a trigonometric "core".  We will also introduce a lot of standard terminology that is used to describe such trigonometric graphs.  This terminology will help us to recognize trigonometric functions in different applied settings.

Describing Trigonometric Graphs

Using our transformational graphing technique, we can quickly sketch the graph of a function like f(x) = -2sin(3(x - p/4)) + 1:

Corresponding
Algebraic formula
sin(x) sin(3x) sin(3(x - p/4))   2sin(3(x - p /4)) -2sin(3(x - p /4)) -2sin(3(x - p/4)) + 1
Geometric Effect Take the graph of the
sin function
Shrink horizontally
by a factor of 1/3
Shift right
p/4 units
Stretch vertically
by a factor of 2
Flip
vertically
Shift up
1 unit
Numerical Effect   Divide by 3 Add p/4 Apply the sin
function
Multiply by 2 Negate Add 1
Numerical 
Results
Inputs Outputs
0 0 p/4 0 0 0 1
p/2 p/6 5p/12 1 2 -2 -1
p p/3 7p/12 0 0 0 1
3p/2 p/2 3p/4 -1 -2 2 3
2p 2p/3 11p/12 0 0 0 1

This gives the following table of values:

t -2sin(3(x - p/4)) + 1
p/4
5p/12
7p/12
3p/4
11p/12
1
-1
1
3
1

with the graph:

Even more so than in other types of graphs, such as linear and exponential graphs, the vertical scaling, vertical shift, and horizontal scaling are clearly visible in this graph.  For example, the vertical shift shows up if we focus on the points that were along the x-axis in the original sin graph.  The line through these points:

is sometimes called the baseline of the sin graph, and we say that the graph has undergone a "baseline shift" of 1 unit up.  Notice how this effectively describes the "center" of the graph in the vertical direction.  Just as your "baseline" heart rate might be 72 beats/min, and will go up or down depending on your level of exertion, the baseline of a trigonometric function is the "average" level of the function from which it varies up or down.

Likewise, the vertical scaling factor is clearly visible as the distance from the baseline to the highest or lowest points on the graph.  This is generally referred to as the amplitude of the graph:

We can see this has an amplitude of 2.  Just as the words "ample" and "enough" are synonymous, the Merriam-Webster dictionary gives "fullness" as a synonym for "amplitude"; in general, the term connotes the "size" of the function.

Finally, the horizontal scaling is visible by looking at the distance between consecutive highest or lowest points of the graph.  This distance is referred to as the wavelength of the graph; if we view the graph as a picture of water waves, this is literally the distance between two consecutive "waves":

Note: This is exactly the same as its period.  In this case, we can see that the wavelength is 11/12 p - 1/4 p = 2p/3.  Since we know that the natural period of the sin function is 2p, and this is a sin graph that has been scaled horizontally by 1/3, this makes sense.  Notice that our method gives us five equally-spaced points to plot over one period of the function; this is sufficient, since we can then simply repeat the pattern in both directions.

Since we know how the vertical scaling, vertical shift, and horizontal scaling show up in the formula, we can identify the amplitude, baseline shift, and wavelength algebraically, as well:

Any function of the form f(x) = ±a sin(±c(x + d)) + b (or ±a cos(±c(x + d)) + b), where a > 0 and c > 0, has amplitude, a, baseline shift, b, and wavelength, 2p/c.

Notice that we have omitted any discussion of the horizontal shift or flips in either direction.  While physicists often refer to these effects as a "phase" shift, because of the symmetries of trigonometric functions, this concept is somewhat ambiguous and hard to define, so we will not use this terminology.

Using this terminology, it becomes easy to recognize how to use trigonometric functions to model different situations.  For example, if we measure the average outside temperature, T (measured in ºF) as a function of the time, t (measured in months, starting in August), we might obtain values like:

t T
0
3
6
9
12
85
50
15
50
85

That is, it is warmest in August, cools off through the Fall, until it is coldest in February, warming back up in the Spring.  We can recognize that the temperature seems to be a periodic function of the time, T = h(t), with period equal to 12, amplitude of 35, and baseline of 50 (i.e., oscillating between a maximum of 85 and a minimum of 15).  Since it starts at its maximum value, this looks like it is related to a cos function, so that we could estimate that h(t) = 35cos((2p/12)t) + 50 = 35cos(pt/6) + 50.  Notice that, using the formula given above, this does have period 2p/(p/6) = 12.

In general, to determine a trigonometric model, f, for a periodic function that starts at its maximum, M, minimum, m, or baseline value, b:

  1. First, determine the "core" function, g:

    1. If it starts at its maximum value, g = cos.

    2. If it starts at its minimum value, g = -cos.

    3. If it starts at its baseline value and goes up, g = sin.

    4. If it starts at its baseline value and goes down, g = -sin.

  2. Determine the amplitude, a, which is half the difference between the maximum and minimum values, i.e., a = (M - m)/2.

  3. Determine the baseline, which is the average of the maximum and minimum values, b = (M + m)/2.

  4. Determine the wavelength, w, which just the period (i.e., distance between repeated maximum or minimum values).

  5. Then we may model the function as f(x) = a g((2p/w)x) + b

For example, it is clear that the height, y, of a Ferris wheel car above the ground, as shown in the following picture:

should be a periodic function of the time, y = f(t).  If we know that it makes one revolution every 15 minutes, then it will have a period of w = 15.  We can see that it will oscillate between a minimum of m = 13 to a maximum of M = 103, from a baseline of b = 58; that is, it has an amplitude of a = 45.  If we assume that, at t = 0, the car starts at the bottom and goes up, then the "core" must be g = -cos, and we can infer that f(t) = -45cos((2p/15)t) + 58.

Practice graphing functions with a "core" of sin or cos, using the terminology of trigonometric functions (i.e., amplitude, baseline shift, and wavelength/period), and determining trigonometric models by completing the following Exercises.


Graphing the Advanced Trigonometric Functions

Our graphing technique works equally well on the advanced trigonometric functions, tan, cot, sec, and csc, once we establish a good collection of points to plot and an accurate picture of their basic graphs.  For example, we have already computed a table of values for sec(x):

x sec(x) = 1/cos(x)
0
p/2
p
3p/2
2p
1/1 = 1
1/0 = undefined
1/(-1) = -1
1/0 = undefined
1/1 = 1

Because this function is undefined at x = p/2 and 3p/2, due to division by 0, it has vertical asymptotes (similar to logarithm graphs) at these points.  Plotting these asymptotes and the other three points gives:

Since a fraction is only 0 when the numerator is 0, we know that sec(x) = 1/cos(x) will never equal 0, that is, the graph will never cross the horizontal axis.  Keeping this in mind, if we "connect-the-dots" in the simplest manner possible, we will obtain a graph like this:

You can compare this with the plot given by XFunctions:

Although XFunctions naturally uses a different plot window than we did and it does not draw in the vertical asymptotes, you can see that our plot accurately represents the true graph.

The basic graph for csc(x) is similar.  Starting from the table:

t csc(t) = 1/sin(t)
0
p/2
p
3p/2
2p
1/0 = undefined
1/1 = 1
1/0 = undefined
1/(-1) = -1
1/0 = undefined

if we sketch the asymptotes at x = 0, p, and 2p, plot the other two points, and connect-the-dots, we obtain the following graph:

Note: While the amplitude for these graphs is ¥ (Why?), we can still see the vertical scaling and period clearly by measuring the vertical and horizontal distances between the "peaks" and "valleys" in the graph.  The baseline is still midway vertically between the "peaks" and "valleys".

The graphs for tan(x) and cot(x) are a bit different.  For example, look at the table of values we computed before:

t tan(t) = sin(t)/cos(t)
0
p/2
p
3p/2
2p
0/1 = 0
1/0 = undefined
0/(-1) = 0
-1/0 = undefined
0/1 = 0

This looks similar to those of the sec and csc functions, in that it shows tan to have an amplitude of ¥.  However, it looks like the actual period of the tan function might be only p.  This is, in fact, the case, as you can see from the plot in XFunctions:

Due to the general shape of the graph, it is rather difficult to see the vertical scaling in the graph without plotting more points. Therefore, when we want to plot tan or cot, we will still find it useful to plot five equally spaced points over one period of the graph, namely:

t cot(t) = cos(t)/sin(t)   t tan(t) = sin(t)/cos(t)
0
p/4
p/2
3p/4
p
undefined
1
0
-1
undefined
and 0
p/4
p/2
3p/4
p
0
1
undefined
-1
0

As with sec and csc, we know that the graphs do not cross the horizontal axis at any other points than those we have listed, so we can  sketch in the asymptotes and connect-the-dots to obtain the graphs:

and

Note: To obtain a graph similar to that in XFunctions, you will need to repeat any graph periodically in both directions.  

Using our transformational graphing technique starting from these points, we can clearly see the vertical scaling, baseline shift, and wavelength.  For example, if we graph g(x) = -2tan(3(x - p/4)) + 1:

Corresponding
Algebraic formula
tan(x) tan(3x) tan(3(x - p/4))   2tan(3(x - p /4)) -2tan(3(x - p /4)) -2tan(3(x - p/4)) + 1
Geometric Effect Take the graph of the
tan function
Shrink horizontally
by a factor of 1/3
Shift right
p/4 units
Stretch vertically
by a factor of 2
Flip
vertically
Shift up
1 unit
Numerical Effect   Divide by 3 Add p/4 Apply the tan
function
Multiply by 2 Negate Add 1
Numerical 
Results
Inputs Outputs
0 0 p/4 0 0 0 1
p/4 p/12 p/3 1 2 -2 -1
p/2 p/6 5p/12 undefined undefined undefined undefined
3p/4 p/4 p/2 -1 -2 2 3
p p/3 7p/12 0 0 0 1

we obtain the following table of values:

t -2tan(3(x - p/4)) + 1
p/4
p/3
5p/12
p/2
7p/12
1
-1
undefined
3
1

with the graph:

Here we have indicated how the graph repeats in both directions.  We can see the baseline at y = 1 and wavelength of 7p/12 - p/4 = p/3.  Although the amplitude is ¥, we can see the vertical scaling of 2, by the vertical distance between the plotted points.

Practice working with these more advanced graphs by plotting functions with tan, cot, sec, or csc as their "core", as in the following Exercises.


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